P10.1 Force and Acceleration AQA GCSE Physics Force And Motion Kerboodle Answers: Page No. 145
1a the resultant force on a sprinterofmass80kg who accelerates at 8m/s2 is
as follows; We know that force = mass*acceleration. Resultant force on
sprinter = 80*8 = 640N.
b acceleration of a car of mass 800 kg acted on by a resultant force of
3200 N is equal to 3200/800 = 4ms -2
2 2a force = mass*acceleration = 0.80*20 = 16N
2 b Mass = force/acceleration = 200/5 = 40 kg.
2 c Acceleration = force/mass = 840/70 = 12m/s 2 .
2 d Force = mass*acceleration = 0.40*6 = 2.4N.
2 e Mass = force/acceleration = 5000/0.20 = 25000kg.
3 a force needed to accelerate the car and the trailer at 2.0 m/s2 is equal to
1500kg × 2m/s 2 = 3000N.
b i Force of the tow bar on the trailer = 300*2 = 600N.
ii Resultant force on the car = 3000N – 600N = 2400N.
4 a the acceleration in the second case was less than before as the total
mass is greater in 2nd case and force is same so acceleration is less.
b calculate the mass of the trolley; F = 0.60 m = 0.48(m + 0.5) gives m =
2.0 kg.
P10.2 Weight and terminal velocity AQA GCSE Physics Force And Motion Kerboodle Answers: Page No. 147
1 a Initial resultant force = weight.
b the weight of the object and the frictional force on it before it
reaches its terminal velocity is frictional force < weight.
c its acceleration after it reaches its terminal velocity is Zero.
d the resultant force on it when it moves at its terminal velocity is Zero.
2 a the weight of a person of mass 5O kg on the Earth Weight =
mass*acceleration = 50*10 = 500N.
b weight of the same person if she was on the Moon is equal to 1.6*50
= 80N.
c A lunar vehicle weighs 300N on the Earth Mass = 300N /10 N/kg = 30 kg.
Weight on Moon = 30 kg × 1.6 N/kg = 48 N.
3 a the parachutist reaches a constant speed because the resultant
force is equal to weight minus frictional force frictional force due to
parachute increases with speed so resultant force on parachutist
decreases, when frictional force = weight, resultant force = 0 and
parachutist moves at terminal velocity.
b i the total weight of the parachutist and the parachute is = 90*10 =
900N.
ii the size and direction of the force of air resistance on the parachute when
the parachutist falls at constant speed is 900 N upwards.
4 a the acceleration of the object in liquid 0.1 Os after it was released.
Gradient measured at 0.10 s = 5.2 m/s 2 .
b the ratio of the drag force to the weight at 0,1 s is about 0.5 because For
mass m, resultant force at 0.10 s = m × acceleration a, a = 5.2 m/s 2 = 0.52g
because resultant force = weight − drag force, drag force = mg − ma= mg −
0.52mg = 0.48mg ≈ half its weight.
P10.3 Forces and braking AQA GCSE Physics Force And Motion Kerboodle Answers: Page No. 149
1 For each of the following factors, identify which distance of a
vehicle (thinking distance or breaking distance) is affected:
a for the road surface braking distance vehicle is affected
b the tiredness of the driver then thinking distance is affected
c poorly maintained brakes then braking distance.
2 a i The thinking distance = 50ft-30ft = 20*0.30m = 6.0m.
ii The braking distance = 125-45 = 80*0.30 = 24.0m.
iii the stopping distance = 175-75 = 100*0.30 = 30.0m.
b A driver has a reaction time of 0.8s. the change in her thinking
distance if she travels at 15m/s instead of 30m/s = (30m/s × 0.8s) −
(15m/s × 0.8s) = 12m.
3 a i the thinking distance of the driver is doubled, assuming that the
driver’s reaction time is unchanged because the Thinking distance
proportional to speed as reaction time is constant.
ii the braking distance is more than doubled as When speed doubled
and braking force constant, braking time greater so braking distance more
than doubles (think about area under velocity–time graph).
b A student thinks that braking distance is proportional to the square
of the speed. Braking distance divided by v 2 same for all three speeds so
braking distance proportional to v 2 so claim is valid.
4 a Deceleration of the vehicle = 312/ 150 = 6.4 m/s 2 .
b Braking force on the car = 1500 kg × 6.4 m/s 2 = 9600 N.
P10.4 Momentum AQA GCSE Physics Force And Motion Kerboodle Answers: Page No. 151
1 a Momentum is the quantity of motion of a moving body, measured as a
product of its mass and velocity. Its unit is momentum =
mass(kg)*velocity(m/s) = kgm/s.
b momentum of 340kg person running at 6n/s, = 40*6 = 240kgm/s.
2 a momentum of an 80kg rugby player running at a velocity of 5 m/s
= 80*5 = 400kgm/s.
b An 800kg car moves with the same momentum as the rugby player
in a. Velocity of the car = 400/800 = 0.5m/s.
c the velocity of a 0.40kg ball that has the same momentum as the
rugby player in a = 400/0.40 = 1000m/s.
3a a the force they exert on each other when they push apart is Equal
and opposite forces.
b the momentum each skater has Just after they separate is equal and
opposite momentum.
c each of their velocities Just after they separate is Velocity of 80 kg
skater = three-quarters velocity of 60 kg skater in opposite direction.
d their total momentum just after they separate, Total momentum = 0.
4 a her momentum = 60*2 = 120 kg m/s.
b the velocity of the other skater is 80v = 60 × 2.0 kg m/s.
v = 60 × 2.0 /80 = 1.5 m/s
P10.5 Using conservation of momentum AQA GCSE Physics Force And Motion Kerboodle Answers: Page No. 153
1 a the momentum of the 1000kg wagon before the collision = 1000*5 =
5000kgm/s.
b the velocity of the wagons after the collision We know that 1000*5+0
= (1000+1500)v, Velocity of the wagon after collision(v) = 5000/2500 =
2m/s.
2 An object is attached to a trolley of mass 0.80 kg, which is then pushed
into an identical stationary trolley at a speed of 1.1 m/s.The two trolleys
couple together and move at a velocity of 0.70 m/s after the collision, the
mass of the object is calculated as follows:
We know that (0.80+m)1.1+0 = (0.80+0.80+m)0.70
0.88+1.1m = 1.12+0.70m
0.4m = 0.24
m = 0.24/0.4 = 0.60kg.
3 a the velocity of the cannon ball when it leaves the cannon calculated as
Mass of canon*speed of canon = – mass of canon ball*velocity of canon
ball= 600*0.5 = – 12*V, V = -25m/s.
b the recoil velocity of the cannon would have been different if a 4kg
cannon ball was used instead, and was fired at the same speed as the 12
kg ball because We know that mv = -MV, 4*-25 = -600V, V = 100/600 =
0.166m/s.
4 Position of right hand block = 2/3*0.6 = 0.4m.
P10.6 Impact forces AQA GCSE Physics Force And Motion Kerboodle Answers: Page No. 155
1 a In a crash test, when a passenger wears a seat belt, it reduces the
impact force on him because it stretches when the car stops moving, so
that the person wearing the belt doesn’t stop immediately.
b A ball of mass 0.12kg moving at a velocity of 18 m/s is caught by a
person in 0.0003 s. Impact force = -0.12*18/0.0003 = -7200N.
2 a An 800kg car travelling at 30m/s is stopped safely when the brakes
are applied.
i Braking force needed to stop it in 6.0s = 800*30/6 = -4000N.
ii Braking force needed to stop it in 6.0s 30s = 800*30/30 = -800N.
b We know that force = mass*change in velocity/time, so force = change in
momentum/time; force is inversely proportional to time so we require much
greater force in short time.
3 a the velocity of the van and the truck immediately after the impact
was 2 m/s as follows: 2000*12 = (2000+10000)V, V = 24000/12000 =
2m/s.
b i Deceleration of the van = 12-2/0.3 = 10/0.3 =33.33m/s.
ii Change of momentum of the van = mass*change in velocity
= 2000(2-12) = -20000kgm/s.
iii Force of impact on the van = -20000/0.3 = -66666.66N.
4 Playgrounds have cushioned surfaces to reduce the risk of injury in
a fall. Impact time on cushioned surface longer than on hard floor, so the
change of momentum per second is less in a fall on cushioned floor than
on hard floor, so impact force is less.
P10.7 Safety first AQA GCSE Physics Force And Motion Kerboodle Answers: Page No. 157
1 a cyclist should always wear a safety helmet when riding a bicycle
because It protects cyclist’s head in collision because when impact occur
helmet increases the time taken to decelerate the head, so it reduces
change of momentum per second and therefore reduces impact of force.
2 a rear-facing car seats are safer than front-facing seats for babies,
because If the car suddenly stopped, the child would press against back of
car seat and therefore spread out the force. The back of the car seat would
prevent child from being thrown forwards.
b an inflated air bag in front of a car user reduces the force on a user
in a head-on crash because An inflated air bag spreads the force of an
impact across the upper part of a person’s body. It also increases the
impact time which reduces the impact force.
3 seat belt reduces the force on a car user when the car suddenly
stops because the time taken to stop someone is longer if they wearing a
seatbelt, so the decelerating force is reduced. The seat belt acts across the
chest, so the force is spread out.
4 a calculate the momentum of the car and the lorry immediately after
the impact, Momentum of the car = 750*9 = 6750kgm/s.
Momentum of the lorry =2150*9 = 19350kgm/s.
P10.8 Forces and elasticity AQA GCSE Physics Force And Motion Kerboodle Answers: Page No. 159
1 a the limit of proportionality of a spring is the limit beyond which
tension no longer proportional to extension.
b the spring constant of a spring is force per unit extension as long as
limit of proportionality not reached .
c the extension of a stretched spring is Increase in length from its
original unstretched length.
2 a a strip of polythene if it is stretched beyond its limit of
proportionality it Does not return to original length when released.
b Rubber band returns to original length when released whereas polythene
strip does not.
3 a i the extension of the spring is 80 mm.
ii the extension of the rubber band is 54 mm.
iii the extension of the polythene strip is 10 mm.
b i extension of the spring when the weight is3.GN is 60 mm.
ii the spring constant of the spring is 3.0 N/ 0.060 m = 50 N/m.
4 a Extension of spring directly proportional to force applied as long as limit
of proportionality not exceeded.
b i force needed to make the spring extend by 0.10m is as follows We
know that Force = spring constant*extension; Force = 25 N/m × 0.10 m =
2.5 N.
ii the extension of the spring when it hangs vertically from a fixed
point and supports a 5.0 N weight at its lower end; Extension = 5/25 =
0.20 m.
Summary questions AQA GCSE Physics Force And Motion: Page No. 160
1 a i the stopping distance of a car is increased if the road is wet
instead of dry because of less friction among tyres and road
ii the stopping distance of a car is increased if the driver is tired
instead of alert because augmented reaction time rises thinking distance
b i the distance travelled by the car in the time it takes the driver to
react is 18 m/s × 0.7 s= 12.6 m
ii stopping distance = 24 m + 12.6 m = 36.6 m
distance from car to dog = 40 m – 36.6 m
= 3.4 m
iii The total mass of the car and its contents is 1200 kg. car’s deceleration
when the brakes are applied, So,
0 = 182 + (2 × a × 24) or − 324 = 48a
a = −328/48 = −6.8 m/s2
braking force = 1200 × (–)6.8 = 8200 N
2 a the weight of the space vehicle on the lunar surface is 200 kg × 1.6
N/kg = 320 N
b the force that each wheel exerts on the lunar surface is 80 N
3 a i the resultant force that produces this acceleration is calculated as 45
kg × 5.0 m/s2 =225 Newton
ii the total weight of the cyclist and the bicycle is 45 kg × 10 N/kg = 450 N
b she can reach a higher speed by crouching than by staying upright
because the she exerts persistent force driving her in forward direction and
crouching by cyclist reduces air resistance which in turn reduces the
frictional force.
4 a the third column of the table is filled as 79, 121, 160, 201, 239
b graph
c If a weight of 7.0N is suspended on the spring, extension of the spring
would be equal to 280 mm
d i spring constant of the spring = 25 N/m
ii An object suspended on the spring gives an extension of 140 mm. Its
weight is 25 N/m × 0.140 m = 3.5 N
5 a i the initial momentum of the car before the brakes are applied is 45
000 kg m/s
ii the braking force is 45 000/12 s = 3750 N
b the momentum of the car changes when the brakes are applied reduces
to zero
c effect on the motion of the car if the brakes had been applied with much
greater force is that it would probably slip and the friction road can apply on
tyres has higher limit
6 a i the momentum of the truck before the collision is 36 000 kg m/s
ii the momentum of the truck; after the collision,20 000 kg m/s
b i the momentum is 16 000 kg m/s
ii the velocity is v = momentum of 1200 kg vehicle after collision/mass of
vehicle = 16 000 kg m/s/1200 kg = 13.3 m/s
7 a Gain of momentum of the football due to the impact. = 0.44 kg × 19
m/s= 8.4 kg m/s
b The impact lasted 0.0384 s Impact force was 8.4 Pa/0.0384 s = 218.75 N
= 220 N (2 s.f.)
Practice questions AQA GCSE Physics Force And Motion:Kerboodle Answer Page No. 161
01.1 reaction time test is better than using the dropping ruler method.
Because it reduces human error/ gives reaction time directly
01.2 independent variable in the test is elapsed time before LED comes
on
01.3 The stopping distance of a car is thinking distance + braking
distance, two factors increase the thinking distance are Tiredness and
alcohol
01.4 The acceleration of a car on a straight road is 3.28 m/s2.The
mass of the car is 1200 kg, resultant force on the car is 3900 N
01.5 The greater the braking force on a car the greater the deceleration of
the car
01.6 the dangers of a very large deceleration on the motion of a car is
that the brakes overheat which result in loss of control of car (skid)
02.1 the factors that should remain constant (the control variables)
during the investigation are surface and weight of trolley
02.2 graph of the results.
02.3 F = 300 N and a = 0.65 m/s2. possible sources of error are, first is
incorrect measurement of time, or trolley placed at incorrect start position
02.4 The student predicted that if the force was doubled the acceleration
would also be doubled. Use gradient of the line on graph to show that value
is 2 or pick a point on the force axis and note the corresponding
acceleration. Double that value for force, and see if the acceleration value
is doubled too
02.5 A stopwatch was used to measure the time taken for the trolley to
travel between points A and B. Another student suggested that tight gates
attached to a data logger would improve the investigation. Advantages of
using light gates and a data logger to measure the acceleration are that
they can store results and less chance of incorrect readings
03.1 The spring is tested in the laboratory. When a force of 10 N is
suspended from the spring, the extension of the spring is 0.02 m. the
spring constant of the spring and give the unit is 500 N/m
03.2 The mass of the baby is 9 kg. the weight of the baby,
gravitational field strength = 9.8 N/kg is 88.2 N
03.3 The baby is placed unsuspended in the harness. His feet are
0.68m from the lower end of the spring. the height above the floor the
lower end of the stretched spring needs be if his feet are just to touch
the floor is 0.86 m (0.68 +0.18)
03.4 The maximum permissible mass of the baby on the bouncer is 15
kg. It is important to test the spring beyond this maximum figure
because to make sure spring does not break or to make sure spring does
not exceed limit of proportionality when used
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This page contains the detailed and easy notes for AQA GCSE Physics Forces for revision and understanding Forces.
AQA GCSE Paper 1: Complete Revision Summary
Banner 1
FORCES
4.5 FORCES
- Scalar and vector Quantites
- Contact and non – contact Forces
- Gravity
- Resultant Forces
- Work Done and Energy transfer
- Forces and Elasticity
- Moments, Levers and Gears
- Pressure in a Fluid
- Atmospheric pressure
Banner 2
FORCES AND MOTION
Distance and Displacement Speed, Velocity, Acceleration Distance Time Graph Velocity Time Graph
NEWTONS LAW OF MOTION
Newtons First Law
Newtons Second Law
Newtons Third Law
Forces and Breaking
Breaking Distance
Thinking Distance
Reaction Time
MOMENTUM
Momentum
Conservation of Momentum
SCALARS AND VECTORS
Banner 3
DISTANCE AND DISPLACEMENT
- Distance is scalar Quantity
- Displacement is a vector Quantity
- Displacement is speed in a given
direction
CONTACT AND NON CONTACT FORCES
Force is push or pull on an object that causes an object due to
interaction with another object that causes an object to:-
a) change speed
b) Change direction
c) change shape
FRICTION FORCE
It is a necessary evil
Friction is a contact force that opposed motion between the
two surfaces that are in physical contact.
- It is a resistive force
- It happens in the opposite direction of motion.
a) It helps to light a matchstick.
b) The friction between the tyres and the
roads prevent the vehicle to slide.
a) It can cause wear and tear of machines
b) It can cause wear and tear of tyres
NEWTONS THIRD LAW OF MOTION
THE SKATERS MOVE TOWARDS EACH OTHER
AS THEY PULL ON EACH OTHER WITH EQUAL
AND OPPOSITE FORCE
RESULTANT FORCES
Net force = 0 N ( forces are balanced)
The body will stay at rest
It is the total force that acts on the body. It is the sum
of all the forces that acts on the body .
The resultant force decides the speed and the direction
of the body.
BALANCED
If the resultant force is zero
- If the forces are balanced and
the body is at rest then it will
stay at rest. - If the forces are balanced and the
body is moving it will keep on moving
with the same speed and direction
UNBALANCED
If the resultant force is
non zero
The body will
move in the direction
of resultant force.
Banner 4
NEWTONS FIRST LAW OF MOTION
If an object is at rest it will remain at rest
If an object is in motion it will continue to move with the same
speed and direction unless no resultant force acts on it.
If the resultant force is non zero or unbalance the object will
move or change speed or direction.
MOMENTS
It is the turning effect of force.
MOMENT =. Force. X Perpendicular distance
from the pivot
Nm = Nxm
Greater distance from the pivot increases
the moment or the turning effort
so a small effor can lift a heavy load.
MOMENT QUESTIONS
Moment = Force X distance
= 30 X 10
= 300 Nm
Q2 The moment of a spanner is 50 Nm.
Calculate the force acting at a distance of
10m from the pivot.
LEVERS AND GEARS
Simple Lever and Force Multipliers
In all these levers, the turning effect of force is greater by increasing the distance of effort
further away from the pivot. It increases the turning effect and multiply the force with a
small effort.
Banner 5
GEARS
GEARS TRANSMIT TURNING EFFECT OF
FORCE
- Gear A Connected to the engine
- Smaller radius
GearB
- Bigger radius
- Connected to Wheels
At the Point of Contact they exert equal and opposite force.
Turning effect is greater due to greater radius .
Low gear
- Low speed and high turning
effect - small gear wheel run a bigger
gear wheel - bigger wheel has greater turning
effect but slow speed
High Gear
- High speed and
low turning effect - large gear wheel run
a smaller gear wheel - small gear wheel force
acts near to the shaft it
run faster with a high speed
but lower turning effect
Baneer 6
MOMENTS AND EQUILBRIUM
The sum of clockwise moments=
The sum of anticlockwise moments
Banner 7
WEIGHT OF THE BEAM
CENTRE OF MASS
It is the point at which the entire mass of the object
can be thought as being concentrated.
CENTRE OF MASS FOR SYMMETRIC OBJECTS
- It is along the point of symmetry
- If there more line of symmetry the centre of mass
is at the intersection of lines of symmetry
SUSPENDED OBJECT
- The center of mass is directly below the point of suspension.
- When suspended, the weight will give the turning effect and
it will come back to its equilbrium position
Banner 8
SPEED, VELOCITY AND ACCELERATION
Q1 A car is travelling at the speed of 20 m/s.
Calculate the distance covered in 10 minutes
DISTANCE TIME GRAPHS
VELOCITY TIME GRAPHS
SLOPE OF VELOCITY TIME GRAPH= ACCELERATION
AREA UNDER THE GRAPH = DISTANCE
Banner 9
EQUATIONS OF MOTION
NEWTONS SECOND LAW OF MOTION
The acceleration of a body is
a) directly proportional to the resultant force
b) inversely proportional to the mass of an object
NEWTONS SECOND LAW OF MOTION
F = Mass x acceleration
SPEED UP
- The velocity of the object increases.
- The object accelerated.
- The resultant force is in the direction of motion.
SPEED DOWN
- The velocity of the object decreases .
- The object is decelerated.
- The resultant force is opposite to the direction of motion.
Banner 10
WEIGHT AND TERMINAL VELOCITY
Weight (N) = Mass (Kg) x GRAVITATIONAL FIELD
STRENGTH (N/Kg)
g on earth = 9.8 N/Kg
or 10 N/Kg
FORCE AND WORK RELATIONSHIP
Q1 Calculate the work done when the force of 100 N moves the object
to a distance of 2m ?
W =F x S
= 100 x 2
= 200J
Q2 Calculate the force applied when 100 J of work is done to move
an object to a distance of 5 m ?
TERMINAL VELOCITY
It is the constant velocity of an object when
the resultant force is zero and the weight
of the body is balanced by the drag and
body has zero acceleration.
Terminal Velocity
TERMINAL VELOCITY IN FLUIDS
- In fluids, weight of the object is balanced by frictional force acting
upwards. - The body falls with constant velocity as net force or resultant force
acting on the object is zero, so the body falls at constant velocity
called the terminal velocity.
FORCES AND BREAKING
STOPPING DISTANCE
- The shortest distance a vehicle
can safely stop in. - It is the sum of thinking distance
and braking distance.
THINKING DISTANCE
The distance travelled by the body
during its reaction time.
= Speed x reaction time
Affected by tiredness, drug, alcohols
as all these affects the reaction time.
BRAKING DISTANCE
- Distance travelled by the body when
the braking force is applied. - Poor weather conditions, road conditions,
poorly maintained vehicles, speed of
the vehicle and the mass of the vehicle
affects the same.
MOMENTUM
Principle of conservation of momentum
- Momentum is a vector quantity .
- It has a magnitude as well as direction.
- Greater the speed, greater the momentum.
- Higher the velocity, higher the momentum.
CONSERVATION OF MOMENTUM
IMPACT FORCES
Relationship between force and momentum
If we increase the time and the momentum is conserved,
the impact force can be decreased.
So greater impact time = reduced impact force
CAR SAFETY FEATURES
All these features increases the impact time, decreasing the momentum and thus
reduced the impact force.
FORCES AND ELASTICITY
Effect of force on elastic objects:-
Change shape or deformation by
- Bending
- Stretching
- Compressing
Elastic Deformation
Object regains its original shape when
the force is removed like stretched rubber
band.
Inelastic Deformation
Object that does not gain its original shape
and changes shape permanently.
Example: overly stretched rubber.
Banner 11
FORCES AND ELASTICITY
HOOKE’s LAW
Force on a spring is directly proportional
to the extension until it reaches its
limit of proportionality.
FORCES AND ELASTICITY
SPRING CONTSTANT
It is the measure of the stiffness of the spring.
Greater the spring constant stiffer is the object.
LIMIT OF PROPORTIONALITY
It is the point upto which the springs obeys
Hooke’s law. Beyond this point, the object
comes in the plastic region and no longer
obeys the Hooke’s law.
Q1 Calculate the force applied on the spring when
it is extended by 2m. The spring constant is 5N/m
F = k e
=5 x 2
=10 N
Q2 Calculate the spring constant of a spring when a force
of 50N extends the spring by 5 m.
K F/e
=50/2
=10 N/m
ELASTIC POTENTIAL ENERGY
E =1/2 k e 2
E = Elastic potential energy (J)
K = spring constant (N/m)
e= extention in the spring (m)
Elastic potential energy is the
energy stored in the spring when
it is stretched or compressed.
PRESSURE
- Greater the force greater the pressure .
- Smaller the area more will be the pressure exerted.
PRESSURE IN LIQUIDS
ATMOSPHERIC PRESSURE
- Atmospheric pressure arises due to
collission of the airparticles with the
Earth’s surface. - It decreases with altitude as the number
of particles decreases with height
causing decrease in weight.
Banner 12
KEY TERMS
a) Forces :-Force is push or pull on an object that causes an
object due to
interaction with another object that causes an
object to:-
a) change speed
b) Change direction
c) change shape
b) Scalar :-Quantity that has magnitude only.
eg Length, Area, Volume etc .
c) Vector : Quantity that has magnitude
as well as direction. eg Displacement,
velocity, acceleration, momentum.
d) Friction :Friction is a contact force that opposed motion
between the
two surfaces that are in physical contact.
e) Newton First Law of Motion :-If an object is at rest it will remain at rest
If an object is in motion it will continue to
move with the same
speed and direction unless no resultant force
acts on it.
f) Newton Second Law of Motion:- The acceleration of a body is
a) directly proportional to the resultant force
b) inversely proportional to the mass of an object.
g) Newton Third Law of Motion:-For an every action force, there is an
equal and opposite reaction force.
h) Resultant Force It is the total force that
acts on the body. It is the sum
of all the forces that acts on the body .The
resultant force decides the speed and the
direction
of the body.
i) Free Body digrams are the graphical
illustration to represent all the forces
acting on a body.
j) Moments: It is the turning effect of force. It
is calculated by force multiplied by the
perpendicular distance from the pivot.
k) Levers : In all these levers, the turning
effect of force is greater by increasing the
distance of effort
further away from the pivot. It increases the
turning effect and multiply the force with a
small effort.
l) Gears : GEARS TRANSMIT TURNING EFFECT OF
FORCE.
m)Centre of Mass: It is the point at which the
entire mass of the object
can be thought as being concentrated.
n) Speed: It is the distance travelled divided by
the time taken.
o) Velocity: Speed in a given direction.
p) Acceleration: It is the change in speed over
time taken.
q)Weight: It is the force acting on the body due to gravity.
r) Terminal Velocity: It is the constant velocity
of an object when
the resultant force is zero and the weight
of the body is balanced by the drag and
body has zero acceleration.
s) Thinking Distance: The distance travelled by
the body during its reaction time.
t) Breaking Distance: Distance travelled by the
body when the braking force is applied.
u)Momentum: It is the product of mass and
velocity.
v) Conservation of Momentum: In a closed system,
the momentum before
the collission and after the collission remains
unchanged.
w) Hooke’s Law: Force on a spring is directly
proportional
to the extension until it reaches its
limit of proportionality.
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References:
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Make sure you have watched the above videos and are familiar with the key definations before trying these questions. It is also good to time yourself while doing these questions so that you can work on the speed as well.